CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

A train travels at a certain average speed of a distance of $$63\ km$$ and then travels at a distance of $$72\ km$$ at an average speed of $$6\ km/hr$$   more than its original speeds. If it takes $$3$$ hours to complete total journey, what is the original average speed?


Solution

Let, the original average speed be $$x$$ km/hr.
Time taken at a speed of $$x$$ km/hr $$(t_1)=\dfrac{63}{x}$$ hrs
Time taken at a speed of $$x+6$$ km/hr $$(t_2)=\dfrac{72}{x+6}$$ hrs

$$t_1+t_2=3$$

$$\dfrac{63}{x}+\dfrac{72}{x+6}=3$$

$$\implies \dfrac{63((x+6)+72x}{x(x+6)}=3$$

$$\implies 63x+378+72x=3x(x+6)$$

$$\implies 378+135x=3x^2+18x$$

$$\implies 3x^2+18x-135x-378=0$$

$$\implies 3x^2-117x-378=0$$

$$\implies 3(x^2-39x-126)=0$$

$$\implies x^2-39x-126=0$$

$$\implies x^2-42x+3x-126=0$$

$$\implies x(x-42)+3(x-42)=0$$

$$\implies (x-42)(x+3)=0$$

Either $$x=42$$ or $$x=-3$$

$$\because$$ speed cannot be negative. $$x=42$$ is considered.

$$\therefore$$ The original average speed is $$42$$ km/hr

Physics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image